Answer:
12 m/s
Explanation:
From the question,
Applying the law of conservation of momentum,
total momentum before collision = Total momentum after collision
mu+Mu' = mv+Mv'........................... Equation 1
Where m = mass of the bullet, u = initial velocity of the bullet, M = combined mass of the gun and the puck, u' = initial velocity of the gun and the puck, v = final velocity of the bullet, v' = final velocity of the gun and the puck
make v the subeject of the equation
v = [(mu+Mu')-Mv']/m................. Equation 2
Given: m = 5.00 g = 0.005 kg, M = 120 g = 0.12 kg, u = u' = 0 m/s (at rest), v' = 0.5 m/s
Substitute these values into equation 2
v = [0-(0.12×0.5)]/0.005
v = -0.06/0.005
v = -12 m/s
The negative sign can be ignored since we are looking for the speed, which has only magnitude.
Hence the speed of the bullet is 12 m/s
